JEE Main & Advanced Physics Electro Magnetic Induction Motional Emi in Loop by Generated Area

If conducting rod moves on two parallel conducting rails as shown in following figure then phenomenon of induced emf can also be understand by the concept of generated area (The area swept of conductor in magnetic field, during it's motion)

As shown in figure in time t distance travelled by conductor = vt

Area generated \[A=lvt\]. Flux linked with this area \[\phi =BA=Blvt\]. Hence induced emf \[|e|\,=\frac{d\varphi }{dt}=Bvl\]

(1) Induced current : \[i=\frac{e}{R}\]\[=\frac{Bvl}{R}\]

(2) Magnetic force : Conductor PQ experiences a magnetic force in opposite direction of it's motion and \[{{F}_{m}}=Bil=B\left( \frac{Bvl}{R} \right)\,l\]\[=\frac{{{B}^{2}}v{{l}^{2}}}{R}\]

(3) Power dissipated in moving the conductor : For uniform motion of rod PQ, the rate of doing mechanical work by external agent or mech. Power delivered by external source is given as \[{{P}_{mech}}={{P}_{ext}}=\frac{dW}{dt}={{F}_{ext}}.\,v=\frac{{{B}^{2}}v{{l}^{2}}}{R}\times v\]\[=\frac{{{B}^{2}}{{v}^{2}}{{l}^{2}}}{R}\]

(4) Electrical power : Also electrical power dissipated in resistance or rate of heat dissipation across resistance is given as \[{{P}_{thermal}}=\frac{H}{t}={{i}^{2}}R={{\left( \frac{Bvl}{R} \right)}^{2}}.R\];\[{{P}_{thermal}}=\frac{{{B}^{2}}{{v}^{2}}{{l}^{2}}}{R}\]

(It is clear that \[{{P}_{mech.}}={{P}_{thermal}}\] which is consistent with the principle of conservation of energy.)

(5) Motion of conductor rod in a vertical plane : If conducting rod released from rest (at \[t=0\]) as shown in figure then with rise in it's speed (v), induces emf (e), induced current (i), magnetic force \[({{F}_{m}})\] increases but it's weight remains constant.

Rod will achieve a constant maximum (terminal) velocity \[{{v}_{T}}\] if \[{{F}_{m}}=mg\] So   \[\frac{{{B}^{2}}v_{T}^{2}{{l}^{2}}}{R}=mg\]    \[\Rightarrow \] \[{{v}_{T}}=\frac{mgR}{{{B}^{2}}{{l}^{2}}}\]

Special cases Motion of train and aeroplane in earth's magnetic field

Induced emf across the axle of the wheels of the train and it is across the tips of the wing of the aeroplane is given by \[e={{B}_{v}}lv\] where \[l=\] length of the axle or distance between the tips of the wings of plane, \[{{B}_{v}}=\] vertical component of earth's magnetic field and v = speed of train or plane.